# Probability of x pocket pairs at a table of n people (NLHE)?

With n people at a table, what is the probability that x of them are dealt pocket pairs? There are several easy ways to approximate this but I was wondering there was an elegant solution. Any takers?

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Just from the top of my head I would say: I think Radu Murea's calculations are pretty ok, however I think that it really is mandatory to calculate that someone else can receive the 2nd card that could make your pair. You could do it in 2 steps:

• What is the chance to receive a pocket if you were dealt 2 cards after each other:

First we receive any card (let's say a 6). Then, what is the chance to receive another 6 from that deck? 3/51 since there are 3 sixes left and 51 cards in the deck. If you were playing with n players, that would be n times 3/51 (1/17). Let's say, you play with 7 people, that would be 7/17 chance. However, we need to elliminate the fact that other people could get your cards as well...

• Chance that someone else gets your card:

If you play with n players (let's say 5), there would be a chance first there would be 3/51, then 3/50, then 3/49 and then 3/48 that one of the players would get your card. Once you receive your second card, you would have 3/47 chance (if your card wasn't gone yet) that you would receive a pocket.

So, calculating both together, I would say that the chance of getting a pocket pair with n players is around: n/17 - n * 3/(51 - n/2 + 1)

I am of course not sure if this is correct. It's just something I came up with in my head so...

EDIT: I am 100% sure my calculations are incorrect. Maybe someone can finetune this. I will also think about it with pen and paper.

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I assume the value `n` is reasonable for a poker table (n = [2, 10]). In this case, it won't be relevant for the result.

You give the first player a card. The probability that the second card you give him has the same rank so that he'll get a pocket pair is: `3 / 51` (the 3 cards of the same rank that remained out of the remaining total of 51 cards).

After this, you can repeat the same rationale for the 2nd player: first cards and `3 / 49` chance that the second one will be a matching rank.

Extrapolating to `x` players, you get the following formula (in pseudo-code):

``````probability = 1
for i = 1 to x
probability = probability * (3 / (52 - 2 * i + 1))
``````

Like I said: `n` is irrelevant in this calculation. And this makes sense, if you think about it.

The problem with this formula is that it doesn't take into account the situation when the matching card of the one you get was already dealt to another player. But there's just no way to know that... (or maybe there is, but it's way too complicated).

PS: Applying the formula to 3 players gives you `1 / 4350`. It's not that improbable. Actually, it's sufficiently probable that it made this actually happen.

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Awesome link :)) Thanks! – Roman Mik Aug 15 '14 at 19:55